Categorical Deduction for 64-Sq. Diagrams
Note: this paper shows the WRONG METHOD, although with a lot of promise. For the real method, see: https://www.academia.edu/10161352/Categorical_Deduction_for_64_Categories
[THIS METHOD REMAINS INCOMPLETE! (HERE BUT NOT THERE)]
Recall that the method for 16-SQUARE DIAGRAMS
Involved FOUR DEDUCTIONS
Those deductions were:
[A] ABFE-CDHG-KOPL-IMNJ
[B] ABFE-CGHD-KOPL-IJNM
[C] AEFB-CDHG-KLPO-IMNJ
[D] AEFB-CGHD-KLPO-IJNM
Each of the letters shown in the
previous diagram
DCBA
GHFE…
LKJI
PONM
Now refers to four squares in the new diagram.
The new diagram has:
A first row:
8,7,6,5,4,3,2,1
A second row: 16,15,
14,13,12,11,10,9
A third row: 24,23,
22,21,20,19,18,17
A fourth row: 32, 31,
30,29,28,27,26,25
A fifth row: 40,39,
38,37,36,35,34,33
A sixth row: 48,47,
46,45,44,43,42,41
A seventh row: 56,55,
54,53,52,51,50,49
And,
An eighth row: 64, 63
62,61,60,59,58,57
Remember, there are other
methods
For showing the categories
The categories could be shown
cyclically
Within each quadrant for example.
I will keep the method I showed here
Because I feel it is the most objective.
Now remember, each of the
categories from 16 square
Corresponds to four of the categoiries
In 64 squares.
64 = 16 X 4
Otherwise the method would not
Reveal itself
So easily.
Now we know that:
‘A’ refers to
1.2,10,9
[Listed in cyclical order]
‘B’ refers to
3,4, 12,11
‘C’ refers to
5,6,14,13
‘D’ refers to
7,8, 16,15
‘E’ refers to
17,18,26,25
‘F’ refers to
19,20,28,27
‘G’ refers to
21,22,30,29
‘H’ refers to
23,24,32,31
‘I’ refers to
33,34,42,41
‘J’ refers to
35,36,44,43
‘K’ refers to
37,38,46,45
‘L’ refers to
39,40,48,47
‘M’ refers to
49,50,58,57
‘N’ refers to
51,52,60,59
‘O’ refers to
53,54,62,61
‘P’ refers to
55,56,64,63
Now, we know that opposite
numbers do not combine!
In terms of coherent quadra,
That means position A
Does not go with position C
Position B
Does not go with position D
That process was used once in
The 16-SQ diagram
Now we apply it again in the
64-SQ diagram
AGAIN,
The deductions for 16-SQ.
WERE:
[A] ABFE-CDHG-KOPL-IMNJ
[B] ABFE-CGHD-KOPL-IJNM
[C] AEFB-CDHG-KLPO-IMNJ
[D] AEFB-CGHD-KLPO-IJNM
So, the deductions for 64-SQ.
Merely involve:
Applying the cyclic order
For each quadrant
(within the numbers…)
SUCH THAT:
The two combinations for
Each cycle
Are maintained AT
EVERY SET LEVEL
Before we ascertain that,
We must find the quadra for every level.
The first two quadrant levels refer to the
16-SQ. diagram.
The third quadrant level refers to the
numbers.
We have already determined
the corresponding numbers
Quadrant A refers to:
ABFE
Quadrant B refers to:
CDHG
Quadrant C refers to:
KLPO
Quadrant D refers to:
IJNM
Now we substitute the
numbers:
QUADRANT A =
A: [1.2,10,9]
B: [3,4, 12,11]
F: [19,20,28,27]
E: [17,18,26,25]
QUADRANT B =
C: [5,6,14,13]
D: [7,8, 16,15]
H: [23,24,32,31]
G: [21,22,30,29]
QUADRANT C =
K: [37,38,46,45]
L: [39,40,48,47]
P: [55,56,64,63]
O: [53,54,62,61]
QUADRANT D =
I: [33,34,42,41]
J: [35,36,44,43]
N: [51,52,60,59]
M: [49,50,58,57]
NOW,
Part A of every level
Only relates with part B,D
Of every level
Part B of every level
Only relates with part C,A
Of every level
Part C of every level
Only relates with part D,A
Of every level
Part D of every level
Only relates with part A,C
Of every level
THEREFORE,
Nothing from quadrant A
Relates with quadrant C
Nothing from quadrant B
Relates with quadrant D
And vice versa
This includes the sections of
The numbers which
Correspond to those quadra
At the third set level.
Therefore, we take the 16-Sq.
Deductions:
[A] ABFE-CDHG-KOPL-IMNJ
[B] ABFE-CGHD-KOPL-IJNM
[C] AEFB-CDHG-KLPO-IMNJ
[D] AEFB-CGHD-KLPO-IJNM
The simplest answer is to apply it
To each of the quadra.
This would leave us with eight
deductions, as predicted
Once the deduction is applied
to the overall quadra.
In QUADRANT A:
ABCDEFGH
IJKLMNOP
Refers to:
1,2,3,4
9,10,11,12
17,18,19,20
25,26,27,28
In QUADRANT B:
ABCDEFGH
IJKLMNOP
Refers to:
5,6,7,8
13,14,15,16
21,22,23,24
29,30,31,32
In QUADRANT C:
ABCDEFGH
IJKLMNOP
Refers to:
37,38,39,40
45,46,47,48
53,54,55,56
61,62,63,64
In QUADRANT D:
ABCDEFGH
IJKLMNOP
Refers to:
33,34,35,36
41,42,43,44
49,50,51,52
57,58,59,60
Now a 16-SQ deduction
for Quadrant A
[A] 1,2,10,9-3,4,12,11-
19,27,28,20-17,25,26,18
[B]1,2,10,9-3,11,12,4-
19,27,28,20-17,18,26,25
[C]1,9,10,2-3,4,12,11-
19,20,28,27-17,25,26,18
[D]1,9,10,2- 3,11,12,4-
19,20,28,27-17,18,26,25
Now a 16-SQ Deduction
for Quadrant B
[A] 5,6,14,13-7,8,16,15-
23,31,32,24-21,29,30,22
[B] 5,6,14,13-7,15,16,8-
23,31,32,24-21,22,30,29
[C] 5,13,14,6-7,8,16,15-
23,24,32,31-21,29,30,22
[D] 5,13,14,6-7,15,16,8-
23,24,32,31-21,22,30,29
Now a 16-SQ Deduction
for Quadrant C
[A] 37,38,46,45-39,40,48,47-
55,63,64,56-53,61,62,54
[B] 37,38,46,45-39,47,48,40-
-55,63,64,56-53,54,62,61
[C] 37,45,46,38-39,40,48,47-
-55,56,64,63-53,61,62,54
[D] 37,45,46,38-39,47,48,40-
55,56,64,63-53,54,62,61
Now a 16-SQ Deduction
for Quadrant D
[A] 33,34,42,41-35,36,44,43-
51,59,60,52-49,57,58,50
[B] 33,34,42,41-35,43,44,36-
-51,59,60,52-49,50,58,57
[C] 33,41,42,34-35,36,44,43-
51,52,60,59-49,57,58,50
[D] 33,41,42,34-35,43,44,36-
51,52,60,59-49,50,58,57
We are nearing our final solution!
Now we simply apply the
Formula:
ABCD and ADCB
On two levels!
In every quadrant
It also takes the order:
ADCB.
Thus, Quadrant A
Is not only:
[A] 1,2,10,9-3,4,12,11-
19,27,28,20-17,25,26,18
[B]1,2,10,9-3,11,12,4-
19,27,28,20-17,18,26,25
[C]1,9,10,2-3,4,12,11-
19,20,28,27-17,25,26,18
[D]1,9,10,2- 3,11,12,4-
19,20,28,27-17,18,26,25
But,
[A] 1,2,10,9-3,4,12,11-
19,27,28,20-17,25,26,18
[D]1,9,10,2- 3,11,12,4-
19,20,28,27-17,18,26,25
[C]1,9,10,2-3,4,12,11-
19,20,28,27-17,25,26,18
[B]1,2,10,9-3,11,12,4-
19,27,28,20-17,18,26,25
Quadrant B is not only:
[A] 5,6,14,13-7,8,16,15-
23,31,32,24-21,29,30,22
[B] 5,6,14,13-7,15,16,8-
23,31,32,24-21,22,30,29
[C] 5,13,14,6-7,8,16,15-
23,24,32,31-21,29,30,22
[D] 5,13,14,6-7,15,16,8-
23,24,32,31-21,22,30,29
But,
[A] 5,6,14,13-7,8,16,15-
23,31,32,24-21,29,30,22
[D] 5,13,14,6-7,15,16,8-
23,24,32,31-21,22,30,29
[C] 5,13,14,6-7,8,16,15-
23,24,32,31-21,29,30,22
[B] 5,6,14,13-7,15,16,8-
23,31,32,24-21,22,30,29
Quadrant C is not only:
[A] 37,38,46,45-39,40,48,47-
55,63,64,56-53,61,62,54
[B] 37,38,46,45-39,47,48,40-
-55,63,64,56-53,54,62,61
[C] 37,45,46,38-39,40,48,47-
-55,56,64,63-53,61,62,54
[D] 37,45,46,38-39,47,48,40-
55,56,64,63-53,54,62,61
But,
[A] 37,38,46,45-39,40,48,47-
55,63,64,56-53,61,62,54
[D] 37,45,46,38-39,47,48,40-
55,56,64,63-53,54,62,61
[C] 37,45,46,38-39,40,48,47-
-55,56,64,63-53,61,62,54
[B] 37,38,46,45-39,47,48,40-
-55,63,64,56-53,54,62,61
Quadrant D is not only:
[A] 33,34,42,41-35,36,44,43-
51,59,60,52-49,57,58,50
[B] 33,34,42,41-35,43,44,36-
-51,59,60,52-49,50,58,57
[C] 33,41,42,34-35,36,44,43-
51,52,60,59-49,57,58,50
[D] 33,41,42,34-35,43,44,36-
51,52,60,59-49,50,58,57
But,
[A] 33,34,42,41-35,36,44,43-
51,59,60,52-49,57,58,50
[D] 33,41,42,34-35,43,44,36-
51,52,60,59-49,50,58,57
[C] 33,41,42,34-35,36,44,43-
51,52,60,59-49,57,58,50
[B] 33,34,42,41-35,43,44,36-
-51,59,60,52-49,50,58,57
Thus, the overall set
takes the order
ABCD and ADCB
With alternation within
each category.
According to the above
dualities…
However, B must
remain opposite of D
And A must remain
Opposite of C…
Thus,
With A.A is
C.B…
With A.B is C.A…
With B.A is D.B…
With B.B is D.A…
With C.A is A.B…
With C.B is A.B…
With D.A is B.B…
With D.B is B.A…
3/4ths of these are
superfluous…
Thus, we have the
combinations:
A.A w/ C.B and
A.B w/ C.A and
B.A w/ D.B and
B.B w/ D.A.
The remaining half of
The categories
Are resolved by
the duality
In which A refers
to B or D…
So we have:
A.A (w/ C.B)
=
A.B (w/ C.A)
=
B.A (w/ D.B)
=
AND
B.B (w/ D.A)
=
Thus, the result is actually
equal to
2^2^2 as predicted…
However, notice, that in spite of the two levels of deductions, the 16-deduction level resulted in four SEPARATE DEDUCTIONS for each quadra, which is inadequate.
Note: this paper shows the WRONG METHOD, although with a lot of promise. For the real method, see: https://www.academia.edu/10161352/Categorical_Deduction_for_64_Categories
[THIS METHOD REMAINS INCOMPLETE (HERE, BUT NOT THERE)]
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